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Đáp án:
Giải thích các bước giải:
43.
`n_{H_2}=3,36÷22,4=0,15 mol`
`2Al + 6HCl\to 2AlCl_3 + 3H_2`
0,1 0,3 <— 0,15
`Al_2O_3+ 6HCl \to 2AlCl_3 + H_2O`
`=> m_{Al}=0,1.27=2,7g`
`%mAl=2,7÷7,8.100=34,6%`
`%mAl_2O_3=100%-34,6%=65,4%`
`m_{Al_2O_3}=7,8-2,7=5,1g`
`=> n_{Al_2O_3}=5,1÷102=0,05 mol`
Theo pt:
`n_{HCl}=3nAl+6nAl2O3=3.0,1+6.0,05=0,6 mol`
`=> VddHCl=0,6÷0,5=1,2`
`m_{Muối}=m_{Al}+m_{Cl}`
`=27.(0,1+0,05.2)+35,5.0,6`
`=26,7g`
44.
`Mg + 2HCl\to MgCl_2 + H_2`
`MgO + 2HCl \to MgCl_2 + H_2O`
Đặt `n_{Mg}=a mol`
`n_{MgO}= b mol`
`=> 24a+40b=8,8` (1)
`n_{MgCl_2}=28,5÷95=0,3 mol`
`=> a+b=0,3` (2)
Giải hpt(1)(2)
`=> a=0,2` `b=0,1`
`%m_{Mg}={24.0,2}/8,8}.100=54,54%`
`%m_{MgO}=100%-54,54%=45,46%`
Theo pt: `n_{HCl}=2.0,3=0,6 mol`
`=> C%HCl={0,6.36,5}/{200}.100=10,95%`
45.
`Fe + 2HCl \to FeCl_2 + H_2`
`Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O`
Đặt `n_{Fe}=a mol`
`n_{Fe_2O_3}=b mol`
`=> 56a+160b=13,6` (1)
`n_{HCl}=1.0,5=0,5 mol`
Theo pt:
`n_{HCl}=2a+6b=0,5` (2)
Giải hpt (1)(2)
`=> a=0,1` `b=0,05`
`=> %mFe={56.0,1}/{13,6}.100=41,2%`
`%mFe_2O_3=100%-41,2%=58,8%`
Theo pt: `n_{H_2}=n_{Fe}=0,1 mol`
`=> V=0,1.22,4=2,24l`
`m_{Muối}=m_{Fe}+m_{Cl}`
`=56.(0,1+0,05.2)+35,5.0,5=28,95g`
